# Subspaces of symplectic vector space?

Consider a linear subspace Y of a symplectic vector space (V,w). Its symplectic orthogonal is defined by

$$Y^O=\{v\in V|w(v,u)=0~ \forall ~u \in Y\}$$

Now, apparently, we must have

$$(Y^O)^O=Y$$

for any Y, but this confuses me a bit. Here is what I'm thinking:

Let's say V is for simplicity 6 dimensional and has a basis of unit vectors $$e_1,e_2,e_3,f_1,f_2,f_3$$ such that $$w(e_i,f_j)=\delta_{i,j} ~~,~~ w(e_i,e_j)=w(f_i,f_j)=0$$

Then Y might be e.g. a subspace with basis $$e_1,e_2,f_1$$. Following the definition above, only $$e_3,f_3$$ would have vanishing symplectic product with all three of these vectors, so that $$e_3,f_3$$ are a basis of $$Y^O$$. But then by the same logic $$(Y^O)^O$$ would have a basis $$e_1,e_2,f_1,f_2$$, which is more than the original 3 vectors and cannot be equivalent to Y. So it seems $$Y\neq (Y^O)^O$$ after all.

Is there some mistake in the above logic? How should the argument properly go?

The problem with your example is that you've missed another independent element of $$Y^O$$: $$e_2$$ is also orthogonal to $$Y$$, so that $$e_2, e_3, f_3$$ is a basis of $$(Y^O)^O$$.

I think the statement only holds in the finite-dimensional case, and if we assume that $$V$$ is finite-dimensional then we can use a dimension-counting argument to prove it. Suppose that $$V$$ is $$k$$-dimensional and $$Y$$ is $$n$$-dimensional, and choose a basis $$b_1, \ldots, b_n$$ of Y. Use this to define a map $$\phi : V \to F^n$$ (where $$F$$ is the field over which $$V$$ is defined) by $$\phi(v) = (w(b_1, v), \ldots, w(b_n, v))$$ By the non-degeneracy of $$w$$, this map is surjective. Its kernel is therefore $$k-n$$ dimensional, and is of course equal to $$Y^O$$. Now define a similar map $$\phi^O$$ using a basis for $$Y^O$$, and note that $$Y \subset \ker(\phi^O)$$, and $$\dim(Y) = \dim(\ker(\phi^O))$$. Therefore $$Y = \ker(\phi^O) = (Y^O)^O$$.

There might be a cleaner argument...

• I thought $w(e_2,f_2)\neq 0$, so $f_2$ does not fit the definition of vectors that may be in $Y^O$ while $e_2$ is in Y? Cluld you elaborate a bit on why $f_2$ is allowed within $Y^O$?
– hms
Commented Jul 27, 2019 at 4:25
• I think you misread :-). It is $e_2$, not $f_2$, that is in $Y^O$.
– Rhys
Commented Jul 27, 2019 at 4:26
• Oh, you are right! Thank you so much, that makes a lot of sense!
– hms
Commented Jul 27, 2019 at 4:27
• If $w$ was in fact degenerate, wouldn't $\phi$ still be surjective? $k$ would just include a number of decoupled dimensions?
– hms
Commented Jul 27, 2019 at 4:51
• Do we obtain $dim(ker(\phi^O))$ by direct inspection of an explicit standard (up to isomorphism) symplectic basis, or is there maybe a sleek argument how to get that number?
– hms
Commented Jul 27, 2019 at 4:53